# How do you evaluate [cos(arccos (3/5) - arcsin (4/5))]?

Oct 27, 2015

$\arccos \left(\frac{3}{5}\right) = \arcsin \left(\frac{4}{5}\right)$ so this is just $\cos \left(0\right) = 1$

#### Explanation:

${3}^{2} + {4}^{2} = {5}^{2}$

So a $3 , 4 , 5$ triangle is a right angled triangle.

$\arccos \left(\frac{3}{5}\right) = B = \arcsin \left(\frac{4}{5}\right)$

So: $\cos \left(\arccos \left(\frac{3}{5}\right) - \arcsin \left(\frac{4}{5}\right)\right) = \cos \left(0\right) = 1$